查找数组中零、正数和负数的比率

给定一个大小为N 个整数的整数数组a ，任务是找出数组中最多四位小数的正数、负数和零的比率。

例子：

Input: a[] = {2, -1, 5, 6, 0, -3} 
Output: 0.5000 0.3333 0.1667 
There are 3 positive, 2 negative, and 1 zero. Their ratio would be positive: 3/6 = 0.5000, negative: 2/6 = 0.3333, and zero: 1/6 = 0.1667.

Input: a[] = {4, 0, -2, -9, -7, 1} 
Output: 0.3333 0.5000 0.1667 
There are 2 positive, 3 negative, and 1 zero. Their ratio would be positive: 2/6 = 0.3333, negative: 3/6 = 0.5000, and zero: 1/6 = 0.1667. 

方法：

1. 计算数组中正元素的总数。
2. 计算数组中负元素的总数。
3. 计算数组中零元素的总数。
4. 将正元素、负元素和零元素的总数除以数组的大小，得到比率。
5. 打印数组中正、负和零元素的比率，最多四位小数。

下面是上述方法的实现。

C++

// C++ program to find the ratio of positive,
// negative, and zero elements in the array.
#include
using namespace std;
 
 
// Function to find the ratio of
// positive, negative, and zero elements
void positiveNegativeZero(int arr[], int len)
{
    // Initialize the postiveCount, negativeCount, and
    // zeroCountby 0 which will count the total number
    // of positive, negative and zero elements
    float positiveCount = 0;
    float negativeCount = 0;
    float zeroCount = 0;
 
    // Traverse the array and count the total number of
    // positive, negative, and zero elements.
    for (int i = 0; i < len; i++) {
        if (arr[i] > 0) {
            positiveCount++;
        }
        else if (arr[i] < 0) {
            negativeCount++;
        }
        else if (arr[i] == 0) {
            zeroCount++;
        }
    }
 
    // Print the ratio of positive,
    // negative, and zero elements
    // in the array up to four decimal places.
    cout << fixed << setprecision(4) << (positiveCount / len)<<" ";
    cout << fixed << setprecision(4) << (negativeCount / len)<<" ";
    cout << fixed << setprecision(4) << (zeroCount / len);
    cout << endl;
}
 
// Driver Code.
int main()
{
    // Test Case 1:
    int a1[] = { 2, -1, 5, 6, 0, -3 };
    int len=sizeof(a1)/sizeof(a1[0]);
    positiveNegativeZero(a1,len);
 
    // Test Case 2:
    int a2[] = { 4, 0, -2, -9, -7, 1 };
    len=sizeof(a2)/sizeof(a2[0]);
    positiveNegativeZero(a2,len);
}
 
// This code is contributed by chitranayal

Java

// Java program to find the ratio of positive,
// negative, and zero elements in the array.
 
class GFG {
 
    // Function to find the ratio of
    // positive, negative, and zero elements
    static void positiveNegativeZero(int[] arr)
    {
 
        // Store the array length into the variable len.
        int len = arr.length;
 
        // Initialize the postiveCount, negativeCount, and
        // zeroCountby 0 which will count the total number
        // of positive, negative and zero elements
        float positiveCount = 0;
        float negativeCount = 0;
        float zeroCount = 0;
 
        // Traverse the array and count the total number of
        // positive, negative, and zero elements.
        for (int i = 0; i < len; i++) {
            if (arr[i] > 0) {
                positiveCount++;
            }
            else if (arr[i] < 0) {
                negativeCount++;
            }
            else if (arr[i] == 0) {
                zeroCount++;
            }
        }
 
        // Print the ratio of positive,
        // negative, and zero elements
        // in the array up to four decimal places.
        System.out.printf("%1.4f ", positiveCount / len);
        System.out.printf("%1.4f ", negativeCount / len);
        System.out.printf("%1.4f ", zeroCount / len);
        System.out.println();
    }
 
    // Driver Code.
    public static void main(String args[])
    {
 
        // Test Case 1:
        int[] a1 = { 2, -1, 5, 6, 0, -3 };
        positiveNegativeZero(a1);
 
        // Test Case 2:
        int[] a2 = { 4, 0, -2, -9, -7, 1 };
        positiveNegativeZero(a2);
    }
}

Python3

# Python3 program to find the ratio of positive,
# negative, and zero elements in the array.
 
# Function to find the ratio of
# positive, negative, and zero elements
def positiveNegativeZero(arr):
 
    # Store the array length into the variable len.
    length = len(arr);
 
    # Initialize the postiveCount, negativeCount, and
    # zeroCountby 0 which will count the total number
    # of positive, negative and zero elements
    positiveCount = 0;
    negativeCount = 0;
    zeroCount = 0;
 
    # Traverse the array and count the total number of
    # positive, negative, and zero elements.
    for i in range(length):
        if (arr[i] > 0):
            positiveCount += 1;
        elif(arr[i] < 0):
            negativeCount += 1;
        elif(arr[i] == 0):
            zeroCount += 1;
         
    # Print the ratio of positive,
    # negative, and zero elements
    # in the array up to four decimal places.
    print("{0:.4f}".format((positiveCount / length)), end=" ");
    print("%1.4f "%(negativeCount / length), end=" ");
    print("%1.4f "%(zeroCount / length), end=" ");
    print();
 
# Driver Code.
if __name__ == '__main__':
 
    # Test Case 1:
    a1 = [ 2, -1, 5, 6, 0, -3 ];
    positiveNegativeZero(a1);
 
    # Test Case 2:
    a2 = [ 4, 0, -2, -9, -7, 1 ];
    positiveNegativeZero(a2);
     
# This code is contributed by Rajput-Ji

C#

// C# program to find the ratio of positive,
// negative, and zero elements in the array.
using System;
 
class GFG {
  
    // Function to find the ratio of
    // positive, negative, and zero elements
    static void positiveNegativeZero(int[] arr)
    {
  
        // Store the array length into the variable len.
        int len = arr.Length;
  
        // Initialize the postiveCount, negativeCount, and
        // zeroCountby 0 which will count the total number
        // of positive, negative and zero elements
        float positiveCount = 0;
        float negativeCount = 0;
        float zeroCount = 0;
  
        // Traverse the array and count the total number of
        // positive, negative, and zero elements.
        for (int i = 0; i < len; i++) {
            if (arr[i] > 0) {
                positiveCount++;
            }
            else if (arr[i] < 0) {
                negativeCount++;
            }
            else if (arr[i] == 0) {
                zeroCount++;
            }
        }
  
        // Print the ratio of positive,
        // negative, and zero elements
        // in the array up to four decimal places.
        Console.Write("{0:F4} ", positiveCount / len);
        Console.Write("{0:F4} ", negativeCount / len);
        Console.Write("{0:F4} ", zeroCount / len);
        Console.WriteLine();
    }
  
    // Driver Code.
    public static void Main(String []args)
    {
  
        // Test Case 1:
        int[] a1 = { 2, -1, 5, 6, 0, -3 };
        positiveNegativeZero(a1);
  
        // Test Case 2:
        int[] a2 = { 4, 0, -2, -9, -7, 1 };
        positiveNegativeZero(a2);
    }
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// Javascript program to find the
// ratio of positive, negative, and
// zero elements in the array.
 
// Function to find the ratio of
// positive, negative, and zero elements
function positiveNegativeZero(arr)
{
     
    // Store the array length into
    // the variable len.
    let len = arr.length;
 
    // Initialize the postiveCount,
    // negativeCount, and zeroCountby
    // 0 which will count the total number
    // of positive, negative and zero elements
    let positiveCount = 0;
    let negativeCount = 0;
    let zeroCount = 0;
 
    // Traverse the array and count the
    // total number of positive, negative,
    // and zero elements.
    for(let i = 0; i < len; i++)
    {
        if (arr[i] > 0)
        {
            positiveCount++;
        }
        else if (arr[i] < 0)
        {
            negativeCount++;
        }
        else if (arr[i] == 0)
        {
            zeroCount++;
        }
    }
 
    // Print the ratio of positive,
    // negative, and zero elements
    // in the array up to four decimal places.
    document.write((positiveCount / len).toFixed(4) + " ");
    document.write((negativeCount / len).toFixed(4) + " ");
    document.write((zeroCount / len).toFixed(4));
    document.write("<br>");
}
 
// Driver Code.
 
// Test Case 1:
let a1 = [ 2, -1, 5, 6, 0, -3 ];
positiveNegativeZero(a1);
 
// Test Case 2:
let a2 = [ 4, 0, -2, -9, -7, 1 ];
positiveNegativeZero(a2);
 
// This code is contributed by sravan kumar G
 

输出：

0.5000 0.3333 0.1667 
0.3333 0.5000 0.1667

时间复杂度： O(len)

辅助空间： O(1)

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